7.2 equivalence relationsbtravers.weebly.com/uploads/6/7/2/9/6729909/... · properties of relations...
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§ 7.2 Equivalence Relations
Properties of Relations
DefinitionA relation R : A→ A is said to be reflexive if xRx for all x ∈ A.
ExampleWhich of the following relations are reflexive, where each is definedon the set S = {a, b, c}?
1 R1 = {(a, b), (b, a), (c, a)}2 R2 = {(a, b), (b, b), (b, c), (c, b), (c, c)}3 R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}4 R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}5 R5 = {(a, a), (a, b)}6 R6 = {(a, a), (a, b), (b, b)}
Properties of Relations
DefinitionA relation R : A→ A is said to be reflexive if xRx for all x ∈ A.
ExampleWhich of the following relations are reflexive, where each is definedon the set S = {a, b, c}?
1 R1 = {(a, b), (b, a), (c, a)}
2 R2 = {(a, b), (b, b), (b, c), (c, b), (c, c)}3 R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}4 R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}5 R5 = {(a, a), (a, b)}6 R6 = {(a, a), (a, b), (b, b)}
Properties of Relations
DefinitionA relation R : A→ A is said to be reflexive if xRx for all x ∈ A.
ExampleWhich of the following relations are reflexive, where each is definedon the set S = {a, b, c}?
1 R1 = {(a, b), (b, a), (c, a)}2 R2 = {(a, b), (b, b), (b, c), (c, b), (c, c)}
3 R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}4 R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}5 R5 = {(a, a), (a, b)}6 R6 = {(a, a), (a, b), (b, b)}
Properties of Relations
DefinitionA relation R : A→ A is said to be reflexive if xRx for all x ∈ A.
ExampleWhich of the following relations are reflexive, where each is definedon the set S = {a, b, c}?
1 R1 = {(a, b), (b, a), (c, a)}2 R2 = {(a, b), (b, b), (b, c), (c, b), (c, c)}3 R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}
4 R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}5 R5 = {(a, a), (a, b)}6 R6 = {(a, a), (a, b), (b, b)}
Properties of Relations
DefinitionA relation R : A→ A is said to be reflexive if xRx for all x ∈ A.
ExampleWhich of the following relations are reflexive, where each is definedon the set S = {a, b, c}?
1 R1 = {(a, b), (b, a), (c, a)}2 R2 = {(a, b), (b, b), (b, c), (c, b), (c, c)}3 R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}4 R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}
5 R5 = {(a, a), (a, b)}6 R6 = {(a, a), (a, b), (b, b)}
Properties of Relations
DefinitionA relation R : A→ A is said to be reflexive if xRx for all x ∈ A.
ExampleWhich of the following relations are reflexive, where each is definedon the set S = {a, b, c}?
1 R1 = {(a, b), (b, a), (c, a)}2 R2 = {(a, b), (b, b), (b, c), (c, b), (c, c)}3 R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}4 R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}5 R5 = {(a, a), (a, b)}
6 R6 = {(a, a), (a, b), (b, b)}
Properties of Relations
DefinitionA relation R : A→ A is said to be reflexive if xRx for all x ∈ A.
ExampleWhich of the following relations are reflexive, where each is definedon the set S = {a, b, c}?
1 R1 = {(a, b), (b, a), (c, a)}2 R2 = {(a, b), (b, b), (b, c), (c, b), (c, c)}3 R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}4 R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}5 R5 = {(a, a), (a, b)}6 R6 = {(a, a), (a, b), (b, b)}
Properties of Relations
DefinitionA relation R : A→ A is said to be symmetric if xRy whenever yRx.
ExampleWhich of the following relations are symmetric, where each is definedon the set S = {a, b, c}?
1 R1 = {(a, b), (b, a), (c, a)}2 R2 = {(a, b), (b, b), (b, c), (c, b), (c, c)}3 R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}4 R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}5 R5 = {(a, a), (a, b), (b, a)}6 R6 = {(a, a), (a, b), (b, b)}
Properties of Relations
DefinitionA relation R : A→ A is said to be symmetric if xRy whenever yRx.
ExampleWhich of the following relations are symmetric, where each is definedon the set S = {a, b, c}?
1 R1 = {(a, b), (b, a), (c, a)}
2 R2 = {(a, b), (b, b), (b, c), (c, b), (c, c)}3 R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}4 R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}5 R5 = {(a, a), (a, b), (b, a)}6 R6 = {(a, a), (a, b), (b, b)}
Properties of Relations
DefinitionA relation R : A→ A is said to be symmetric if xRy whenever yRx.
ExampleWhich of the following relations are symmetric, where each is definedon the set S = {a, b, c}?
1 R1 = {(a, b), (b, a), (c, a)}2 R2 = {(a, b), (b, b), (b, c), (c, b), (c, c)}
3 R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}4 R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}5 R5 = {(a, a), (a, b), (b, a)}6 R6 = {(a, a), (a, b), (b, b)}
Properties of Relations
DefinitionA relation R : A→ A is said to be symmetric if xRy whenever yRx.
ExampleWhich of the following relations are symmetric, where each is definedon the set S = {a, b, c}?
1 R1 = {(a, b), (b, a), (c, a)}2 R2 = {(a, b), (b, b), (b, c), (c, b), (c, c)}3 R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}
4 R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}5 R5 = {(a, a), (a, b), (b, a)}6 R6 = {(a, a), (a, b), (b, b)}
Properties of Relations
DefinitionA relation R : A→ A is said to be symmetric if xRy whenever yRx.
ExampleWhich of the following relations are symmetric, where each is definedon the set S = {a, b, c}?
1 R1 = {(a, b), (b, a), (c, a)}2 R2 = {(a, b), (b, b), (b, c), (c, b), (c, c)}3 R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}4 R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}
5 R5 = {(a, a), (a, b), (b, a)}6 R6 = {(a, a), (a, b), (b, b)}
Properties of Relations
DefinitionA relation R : A→ A is said to be symmetric if xRy whenever yRx.
ExampleWhich of the following relations are symmetric, where each is definedon the set S = {a, b, c}?
1 R1 = {(a, b), (b, a), (c, a)}2 R2 = {(a, b), (b, b), (b, c), (c, b), (c, c)}3 R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}4 R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}5 R5 = {(a, a), (a, b), (b, a)}
6 R6 = {(a, a), (a, b), (b, b)}
Properties of Relations
DefinitionA relation R : A→ A is said to be symmetric if xRy whenever yRx.
ExampleWhich of the following relations are symmetric, where each is definedon the set S = {a, b, c}?
1 R1 = {(a, b), (b, a), (c, a)}2 R2 = {(a, b), (b, b), (b, c), (c, b), (c, c)}3 R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}4 R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}5 R5 = {(a, a), (a, b), (b, a)}6 R6 = {(a, a), (a, b), (b, b)}
Properties of Relations
DefinitionA relation R : A→ A is said to be antisymmetric if xRy whenevery 6 Rx.
ExampleWhich of the following relations are antisymmetric, where each isdefined on the set S = {a, b, c}?
1 R1 = {(a, b), (b, a), (c, a), (a, c)}2 R2 = {(a, b), (b, b), (b, c), (a, c)}3 R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}4 R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}5 R5 = {(a, a), (a, b), (b, a)}6 R6 = {(a, a), (a, b), (b, b)}
Properties of Relations
DefinitionA relation R : A→ A is said to be antisymmetric if xRy whenevery 6 Rx.
ExampleWhich of the following relations are antisymmetric, where each isdefined on the set S = {a, b, c}?
1 R1 = {(a, b), (b, a), (c, a), (a, c)}
2 R2 = {(a, b), (b, b), (b, c), (a, c)}3 R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}4 R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}5 R5 = {(a, a), (a, b), (b, a)}6 R6 = {(a, a), (a, b), (b, b)}
Properties of Relations
DefinitionA relation R : A→ A is said to be antisymmetric if xRy whenevery 6 Rx.
ExampleWhich of the following relations are antisymmetric, where each isdefined on the set S = {a, b, c}?
1 R1 = {(a, b), (b, a), (c, a), (a, c)}2 R2 = {(a, b), (b, b), (b, c), (a, c)}
3 R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}4 R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}5 R5 = {(a, a), (a, b), (b, a)}6 R6 = {(a, a), (a, b), (b, b)}
Properties of Relations
DefinitionA relation R : A→ A is said to be antisymmetric if xRy whenevery 6 Rx.
ExampleWhich of the following relations are antisymmetric, where each isdefined on the set S = {a, b, c}?
1 R1 = {(a, b), (b, a), (c, a), (a, c)}2 R2 = {(a, b), (b, b), (b, c), (a, c)}3 R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}
4 R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}5 R5 = {(a, a), (a, b), (b, a)}6 R6 = {(a, a), (a, b), (b, b)}
Properties of Relations
DefinitionA relation R : A→ A is said to be antisymmetric if xRy whenevery 6 Rx.
ExampleWhich of the following relations are antisymmetric, where each isdefined on the set S = {a, b, c}?
1 R1 = {(a, b), (b, a), (c, a), (a, c)}2 R2 = {(a, b), (b, b), (b, c), (a, c)}3 R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}4 R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}
5 R5 = {(a, a), (a, b), (b, a)}6 R6 = {(a, a), (a, b), (b, b)}
Properties of Relations
DefinitionA relation R : A→ A is said to be antisymmetric if xRy whenevery 6 Rx.
ExampleWhich of the following relations are antisymmetric, where each isdefined on the set S = {a, b, c}?
1 R1 = {(a, b), (b, a), (c, a), (a, c)}2 R2 = {(a, b), (b, b), (b, c), (a, c)}3 R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}4 R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}5 R5 = {(a, a), (a, b), (b, a)}
6 R6 = {(a, a), (a, b), (b, b)}
Properties of Relations
DefinitionA relation R : A→ A is said to be antisymmetric if xRy whenevery 6 Rx.
ExampleWhich of the following relations are antisymmetric, where each isdefined on the set S = {a, b, c}?
1 R1 = {(a, b), (b, a), (c, a), (a, c)}2 R2 = {(a, b), (b, b), (b, c), (a, c)}3 R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}4 R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}5 R5 = {(a, a), (a, b), (b, a)}6 R6 = {(a, a), (a, b), (b, b)}
Properties of Relations
DefinitionA relation R : A→ A is said to be transitive if whenever xRy and yRz,xRz.
ExampleWhich of the following relations are transitive, where each is definedon the set S = {a, b, c}?
1 R1 = {(a, b), (b, a), (c, a)}Not transitive because (a, b), (b, a) ∈ R1 but (a, a) 6∈ R1.
2 R2 = {(a, b), (b, b), (b, c), (c, b), (c, c)}Not reflexive because (a, b), (b, c) ∈ R2 but (a, c) 6∈ R2.
Properties of Relations
DefinitionA relation R : A→ A is said to be transitive if whenever xRy and yRz,xRz.
ExampleWhich of the following relations are transitive, where each is definedon the set S = {a, b, c}?
1 R1 = {(a, b), (b, a), (c, a)}
Not transitive because (a, b), (b, a) ∈ R1 but (a, a) 6∈ R1.2 R2 = {(a, b), (b, b), (b, c), (c, b), (c, c)}
Not reflexive because (a, b), (b, c) ∈ R2 but (a, c) 6∈ R2.
Properties of Relations
DefinitionA relation R : A→ A is said to be transitive if whenever xRy and yRz,xRz.
ExampleWhich of the following relations are transitive, where each is definedon the set S = {a, b, c}?
1 R1 = {(a, b), (b, a), (c, a)}Not transitive because (a, b), (b, a) ∈ R1 but (a, a) 6∈ R1.
2 R2 = {(a, b), (b, b), (b, c), (c, b), (c, c)}Not reflexive because (a, b), (b, c) ∈ R2 but (a, c) 6∈ R2.
Properties of Relations
DefinitionA relation R : A→ A is said to be transitive if whenever xRy and yRz,xRz.
ExampleWhich of the following relations are transitive, where each is definedon the set S = {a, b, c}?
1 R1 = {(a, b), (b, a), (c, a)}Not transitive because (a, b), (b, a) ∈ R1 but (a, a) 6∈ R1.
2 R2 = {(a, b), (b, b), (b, c), (c, b), (c, c)}
Not reflexive because (a, b), (b, c) ∈ R2 but (a, c) 6∈ R2.
Properties of Relations
DefinitionA relation R : A→ A is said to be transitive if whenever xRy and yRz,xRz.
ExampleWhich of the following relations are transitive, where each is definedon the set S = {a, b, c}?
1 R1 = {(a, b), (b, a), (c, a)}Not transitive because (a, b), (b, a) ∈ R1 but (a, a) 6∈ R1.
2 R2 = {(a, b), (b, b), (b, c), (c, b), (c, c)}Not reflexive because (a, b), (b, c) ∈ R2 but (a, c) 6∈ R2.
Properties of Relations
Example1 R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}
This is transitive.2 R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}
This is transitive.3 R5 = {(a, a), (a, b)}
This is transitive.4 R6 = {(a, b), (a, c)}
This is transitive.
Properties of Relations
Example1 R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}
This is transitive.
2 R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}This is transitive.
3 R5 = {(a, a), (a, b)}This is transitive.
4 R6 = {(a, b), (a, c)}This is transitive.
Properties of Relations
Example1 R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}
This is transitive.2 R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}
This is transitive.3 R5 = {(a, a), (a, b)}
This is transitive.4 R6 = {(a, b), (a, c)}
This is transitive.
Properties of Relations
Example1 R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}
This is transitive.2 R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}
This is transitive.
3 R5 = {(a, a), (a, b)}This is transitive.
4 R6 = {(a, b), (a, c)}This is transitive.
Properties of Relations
Example1 R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}
This is transitive.2 R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}
This is transitive.3 R5 = {(a, a), (a, b)}
This is transitive.4 R6 = {(a, b), (a, c)}
This is transitive.
Properties of Relations
Example1 R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}
This is transitive.2 R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}
This is transitive.3 R5 = {(a, a), (a, b)}
This is transitive.
4 R6 = {(a, b), (a, c)}This is transitive.
Properties of Relations
Example1 R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}
This is transitive.2 R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}
This is transitive.3 R5 = {(a, a), (a, b)}
This is transitive.4 R6 = {(a, b), (a, c)}
This is transitive.
Properties of Relations
Example1 R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}
This is transitive.2 R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}
This is transitive.3 R5 = {(a, a), (a, b)}
This is transitive.4 R6 = {(a, b), (a, c)}
This is transitive.
Properties of Relations
Example1 R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}
This is transitive.2 R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}
This is transitive.3 R5 = {(a, a), (a, b)}
This is transitive.4 R6 = {(a, b), (a, c)}
This is transitive.
Back To Common Examples
Some Common ExamplesLet Ri : Z→ Z. Determine if each of the following is reflexive,symmetric, antisymmetric, or transitive. Note: a relation could bemore than one of these ...
R1 = {(a, b) | a ≥ b}
R2 = {(a, b) | a < b}R3 = {(a, b) | (a = b) ∨ (a 6= b)}R4 = {(a, b) | a = b + 1}R5 = {(a, b) | a + b ≤ 3}
Back To Common Examples
Some Common ExamplesLet Ri : Z→ Z. Determine if each of the following is reflexive,symmetric, antisymmetric, or transitive. Note: a relation could bemore than one of these ...
R1 = {(a, b) | a ≥ b}R2 = {(a, b) | a < b}
R3 = {(a, b) | (a = b) ∨ (a 6= b)}R4 = {(a, b) | a = b + 1}R5 = {(a, b) | a + b ≤ 3}
Back To Common Examples
Some Common ExamplesLet Ri : Z→ Z. Determine if each of the following is reflexive,symmetric, antisymmetric, or transitive. Note: a relation could bemore than one of these ...
R1 = {(a, b) | a ≥ b}R2 = {(a, b) | a < b}R3 = {(a, b) | (a = b) ∨ (a 6= b)}
R4 = {(a, b) | a = b + 1}R5 = {(a, b) | a + b ≤ 3}
Back To Common Examples
Some Common ExamplesLet Ri : Z→ Z. Determine if each of the following is reflexive,symmetric, antisymmetric, or transitive. Note: a relation could bemore than one of these ...
R1 = {(a, b) | a ≥ b}R2 = {(a, b) | a < b}R3 = {(a, b) | (a = b) ∨ (a 6= b)}R4 = {(a, b) | a = b + 1}
R5 = {(a, b) | a + b ≤ 3}
Back To Common Examples
Some Common ExamplesLet Ri : Z→ Z. Determine if each of the following is reflexive,symmetric, antisymmetric, or transitive. Note: a relation could bemore than one of these ...
R1 = {(a, b) | a ≥ b}R2 = {(a, b) | a < b}R3 = {(a, b) | (a = b) ∨ (a 6= b)}R4 = {(a, b) | a = b + 1}R5 = {(a, b) | a + b ≤ 3}
Equivalence Relations
DefinitionA relation is called an equivalence relation if it is symmetric, reflexiveand transitive.
From our last example, which are equivalence relations?
Common Equivalence Relations
R1 = {(a, b) | a ≥ b}R2 = {(a, b) | a < b}R3 = {(a, b) | (a = b) ∨ (a 6= b)}R4 = {(a, b) | a = b + 1}R5 = {(a, b) | a + b ≤ 3}
Equivalence Relations
DefinitionA relation is called an equivalence relation if it is symmetric, reflexiveand transitive.
From our last example, which are equivalence relations?
Common Equivalence Relations
R1 = {(a, b) | a ≥ b}
R2 = {(a, b) | a < b}R3 = {(a, b) | (a = b) ∨ (a 6= b)}R4 = {(a, b) | a = b + 1}R5 = {(a, b) | a + b ≤ 3}
Equivalence Relations
DefinitionA relation is called an equivalence relation if it is symmetric, reflexiveand transitive.
From our last example, which are equivalence relations?
Common Equivalence Relations
R1 = {(a, b) | a ≥ b}R2 = {(a, b) | a < b}
R3 = {(a, b) | (a = b) ∨ (a 6= b)}R4 = {(a, b) | a = b + 1}R5 = {(a, b) | a + b ≤ 3}
Equivalence Relations
DefinitionA relation is called an equivalence relation if it is symmetric, reflexiveand transitive.
From our last example, which are equivalence relations?
Common Equivalence Relations
R1 = {(a, b) | a ≥ b}R2 = {(a, b) | a < b}R3 = {(a, b) | (a = b) ∨ (a 6= b)}
R4 = {(a, b) | a = b + 1}R5 = {(a, b) | a + b ≤ 3}
Equivalence Relations
DefinitionA relation is called an equivalence relation if it is symmetric, reflexiveand transitive.
From our last example, which are equivalence relations?
Common Equivalence Relations
R1 = {(a, b) | a ≥ b}R2 = {(a, b) | a < b}R3 = {(a, b) | (a = b) ∨ (a 6= b)}R4 = {(a, b) | a = b + 1}
R5 = {(a, b) | a + b ≤ 3}
Equivalence Relations
DefinitionA relation is called an equivalence relation if it is symmetric, reflexiveand transitive.
From our last example, which are equivalence relations?
Common Equivalence Relations
R1 = {(a, b) | a ≥ b}R2 = {(a, b) | a < b}R3 = {(a, b) | (a = b) ∨ (a 6= b)}R4 = {(a, b) | a = b + 1}R5 = {(a, b) | a + b ≤ 3}
An Equivalence Relation Proof
ExampleA relation R is defined on Z by xRy if x + 3y is even. Prove R is anequivalence relation.
Proof.First, we show R is reflexive. Let a ∈ Z. Then a + 3a = 4a = 2(2a)is even since 2a ∈ Z. Therefore, R is reflexive.
An Equivalence Relation Proof
ExampleA relation R is defined on Z by xRy if x + 3y is even. Prove R is anequivalence relation.
Proof.First, we show R is reflexive.
Let a ∈ Z. Then a + 3a = 4a = 2(2a)is even since 2a ∈ Z. Therefore, R is reflexive.
An Equivalence Relation Proof
ExampleA relation R is defined on Z by xRy if x + 3y is even. Prove R is anequivalence relation.
Proof.First, we show R is reflexive. Let a ∈ Z. Then a + 3a = 4a = 2(2a)is even since 2a ∈ Z. Therefore, R is reflexive.
An Equivalence Relation Proof
ExampleA relation R is defined on Z by xRy if x + 3y is even. Prove R is anequivalence relation.
Proof.First, we show R is reflexive. Let a ∈ Z. Then a + 3a = 4a = 2(2a)is even since 2a ∈ Z. Therefore, R is reflexive.
An Equivalence Relation Proof
Proof.We now show R is symmetric.
Assume aRb. Then a + 3b is even. So,there exists k ∈ Z such that a + 3b = 2k, or that a = 2k − 3b.Consider
b + 3a = b + 3(2k − 3b)
= b + 6k − 9b
= 6k − 8b
= 2(3k − 4b)
Since 3k − 4b ∈ Z, b + 3a is even. Therefore, bRa and so R issymmetric.
An Equivalence Relation Proof
Proof.We now show R is symmetric. Assume aRb. Then a + 3b is even.
So,there exists k ∈ Z such that a + 3b = 2k, or that a = 2k − 3b.Consider
b + 3a = b + 3(2k − 3b)
= b + 6k − 9b
= 6k − 8b
= 2(3k − 4b)
Since 3k − 4b ∈ Z, b + 3a is even. Therefore, bRa and so R issymmetric.
An Equivalence Relation Proof
Proof.We now show R is symmetric. Assume aRb. Then a + 3b is even. So,there exists k ∈ Z such that a + 3b = 2k, or that a = 2k − 3b.
Consider
b + 3a = b + 3(2k − 3b)
= b + 6k − 9b
= 6k − 8b
= 2(3k − 4b)
Since 3k − 4b ∈ Z, b + 3a is even. Therefore, bRa and so R issymmetric.
An Equivalence Relation Proof
Proof.We now show R is symmetric. Assume aRb. Then a + 3b is even. So,there exists k ∈ Z such that a + 3b = 2k, or that a = 2k − 3b.Consider
b + 3a = b + 3(2k − 3b)
= b + 6k − 9b
= 6k − 8b
= 2(3k − 4b)
Since 3k − 4b ∈ Z, b + 3a is even. Therefore, bRa and so R issymmetric.
An Equivalence Relation Proof
Proof.We now show R is symmetric. Assume aRb. Then a + 3b is even. So,there exists k ∈ Z such that a + 3b = 2k, or that a = 2k − 3b.Consider
b + 3a = b + 3(2k − 3b)
= b + 6k − 9b
= 6k − 8b
= 2(3k − 4b)
Since 3k − 4b ∈ Z, b + 3a is even. Therefore, bRa and so R issymmetric.
An Equivalence Relation Proof
Proof.We now show R is symmetric. Assume aRb. Then a + 3b is even. So,there exists k ∈ Z such that a + 3b = 2k, or that a = 2k − 3b.Consider
b + 3a = b + 3(2k − 3b)
= b + 6k − 9b
= 6k − 8b
= 2(3k − 4b)
Since 3k − 4b ∈ Z, b + 3a is even. Therefore, bRa and so R issymmetric.
An Equivalence Relation Proof
Proof.We now show R is symmetric. Assume aRb. Then a + 3b is even. So,there exists k ∈ Z such that a + 3b = 2k, or that a = 2k − 3b.Consider
b + 3a = b + 3(2k − 3b)
= b + 6k − 9b
= 6k − 8b
= 2(3k − 4b)
Since 3k − 4b ∈ Z, b + 3a is even. Therefore, bRa and so R issymmetric.
An Equivalence Relation Proof
Proof.We now show R is symmetric. Assume aRb. Then a + 3b is even. So,there exists k ∈ Z such that a + 3b = 2k, or that a = 2k − 3b.Consider
b + 3a = b + 3(2k − 3b)
= b + 6k − 9b
= 6k − 8b
= 2(3k − 4b)
Since 3k − 4b ∈ Z, b + 3a is even.
Therefore, bRa and so R issymmetric.
An Equivalence Relation Proof
Proof.We now show R is symmetric. Assume aRb. Then a + 3b is even. So,there exists k ∈ Z such that a + 3b = 2k, or that a = 2k − 3b.Consider
b + 3a = b + 3(2k − 3b)
= b + 6k − 9b
= 6k − 8b
= 2(3k − 4b)
Since 3k − 4b ∈ Z, b + 3a is even. Therefore, bRa and so R issymmetric.
An Equivalence Relation Proof
Proof.We now show R is transitive.
Assume aRb and bRc. Hence, a + 3band b + 3c are even. So, there exists integers n and m such thata + 3b = 2n and b + 3c = 2m. That is, a = 2n− 3b and3c = 2m− b. Consider
a + 3c = 2n− 3b + 2m− b
= 2n + 2m− 4b
= 2(n + m− 2b)
Since n + m− 2b ∈ Z, it follows that a + 3c is even, and so aRc.Therefore, R is transitive.
Since R is reflexive, symmetric and transitive, R is an equivalencerelation.
An Equivalence Relation Proof
Proof.We now show R is transitive. Assume aRb and bRc.
Hence, a + 3band b + 3c are even. So, there exists integers n and m such thata + 3b = 2n and b + 3c = 2m. That is, a = 2n− 3b and3c = 2m− b. Consider
a + 3c = 2n− 3b + 2m− b
= 2n + 2m− 4b
= 2(n + m− 2b)
Since n + m− 2b ∈ Z, it follows that a + 3c is even, and so aRc.Therefore, R is transitive.
Since R is reflexive, symmetric and transitive, R is an equivalencerelation.
An Equivalence Relation Proof
Proof.We now show R is transitive. Assume aRb and bRc. Hence, a + 3band b + 3c are even.
So, there exists integers n and m such thata + 3b = 2n and b + 3c = 2m. That is, a = 2n− 3b and3c = 2m− b. Consider
a + 3c = 2n− 3b + 2m− b
= 2n + 2m− 4b
= 2(n + m− 2b)
Since n + m− 2b ∈ Z, it follows that a + 3c is even, and so aRc.Therefore, R is transitive.
Since R is reflexive, symmetric and transitive, R is an equivalencerelation.
An Equivalence Relation Proof
Proof.We now show R is transitive. Assume aRb and bRc. Hence, a + 3band b + 3c are even. So, there exists integers n and m such thata + 3b = 2n and b + 3c = 2m. That is, a = 2n− 3b and3c = 2m− b.
Consider
a + 3c = 2n− 3b + 2m− b
= 2n + 2m− 4b
= 2(n + m− 2b)
Since n + m− 2b ∈ Z, it follows that a + 3c is even, and so aRc.Therefore, R is transitive.
Since R is reflexive, symmetric and transitive, R is an equivalencerelation.
An Equivalence Relation Proof
Proof.We now show R is transitive. Assume aRb and bRc. Hence, a + 3band b + 3c are even. So, there exists integers n and m such thata + 3b = 2n and b + 3c = 2m. That is, a = 2n− 3b and3c = 2m− b. Consider
a + 3c = 2n− 3b + 2m− b
= 2n + 2m− 4b
= 2(n + m− 2b)
Since n + m− 2b ∈ Z, it follows that a + 3c is even, and so aRc.Therefore, R is transitive.
Since R is reflexive, symmetric and transitive, R is an equivalencerelation.
An Equivalence Relation Proof
Proof.We now show R is transitive. Assume aRb and bRc. Hence, a + 3band b + 3c are even. So, there exists integers n and m such thata + 3b = 2n and b + 3c = 2m. That is, a = 2n− 3b and3c = 2m− b. Consider
a + 3c = 2n− 3b + 2m− b
= 2n + 2m− 4b
= 2(n + m− 2b)
Since n + m− 2b ∈ Z, it follows that a + 3c is even, and so aRc.Therefore, R is transitive.
Since R is reflexive, symmetric and transitive, R is an equivalencerelation.
An Equivalence Relation Proof
Proof.We now show R is transitive. Assume aRb and bRc. Hence, a + 3band b + 3c are even. So, there exists integers n and m such thata + 3b = 2n and b + 3c = 2m. That is, a = 2n− 3b and3c = 2m− b. Consider
a + 3c = 2n− 3b + 2m− b
= 2n + 2m− 4b
= 2(n + m− 2b)
Since n + m− 2b ∈ Z, it follows that a + 3c is even, and so aRc.Therefore, R is transitive.
Since R is reflexive, symmetric and transitive, R is an equivalencerelation.
An Equivalence Relation Proof
Proof.We now show R is transitive. Assume aRb and bRc. Hence, a + 3band b + 3c are even. So, there exists integers n and m such thata + 3b = 2n and b + 3c = 2m. That is, a = 2n− 3b and3c = 2m− b. Consider
a + 3c = 2n− 3b + 2m− b
= 2n + 2m− 4b
= 2(n + m− 2b)
Since n + m− 2b ∈ Z, it follows that a + 3c is even, and so aRc.Therefore, R is transitive.
Since R is reflexive, symmetric and transitive, R is an equivalencerelation.
An Equivalence Relation Proof
Proof.We now show R is transitive. Assume aRb and bRc. Hence, a + 3band b + 3c are even. So, there exists integers n and m such thata + 3b = 2n and b + 3c = 2m. That is, a = 2n− 3b and3c = 2m− b. Consider
a + 3c = 2n− 3b + 2m− b
= 2n + 2m− 4b
= 2(n + m− 2b)
Since n + m− 2b ∈ Z, it follows that a + 3c is even, and so aRc.Therefore, R is transitive.
Since R is reflexive, symmetric and transitive, R is an equivalencerelation.
Another Example
Example
Let H = {4k | k ∈ Z}.A relation R is defined on Z by aRb ifa− b ∈ H. Show that R is an equivalence relation.
Proof.Reflexive: Let a ∈ Z. Since a− a = 0 = 4 · 0, it follows that aRa.Symmetric: Assume aRb, where a, b ∈ Z. Then, a− b = 4k for somek ∈ Z. Then, b− a = 4(−k), and since −k ∈ Z, it follows thatb− a ∈ H. Therefore, bRa.
Another Example
Example
Let H = {4k | k ∈ Z}.A relation R is defined on Z by aRb ifa− b ∈ H. Show that R is an equivalence relation.
Proof.Reflexive:
Let a ∈ Z. Since a− a = 0 = 4 · 0, it follows that aRa.Symmetric: Assume aRb, where a, b ∈ Z. Then, a− b = 4k for somek ∈ Z. Then, b− a = 4(−k), and since −k ∈ Z, it follows thatb− a ∈ H. Therefore, bRa.
Another Example
Example
Let H = {4k | k ∈ Z}.A relation R is defined on Z by aRb ifa− b ∈ H. Show that R is an equivalence relation.
Proof.Reflexive: Let a ∈ Z. Since a− a = 0 = 4 · 0, it follows that aRa.
Symmetric: Assume aRb, where a, b ∈ Z. Then, a− b = 4k for somek ∈ Z. Then, b− a = 4(−k), and since −k ∈ Z, it follows thatb− a ∈ H. Therefore, bRa.
Another Example
Example
Let H = {4k | k ∈ Z}.A relation R is defined on Z by aRb ifa− b ∈ H. Show that R is an equivalence relation.
Proof.Reflexive: Let a ∈ Z. Since a− a = 0 = 4 · 0, it follows that aRa.Symmetric:
Assume aRb, where a, b ∈ Z. Then, a− b = 4k for somek ∈ Z. Then, b− a = 4(−k), and since −k ∈ Z, it follows thatb− a ∈ H. Therefore, bRa.
Another Example
Example
Let H = {4k | k ∈ Z}.A relation R is defined on Z by aRb ifa− b ∈ H. Show that R is an equivalence relation.
Proof.Reflexive: Let a ∈ Z. Since a− a = 0 = 4 · 0, it follows that aRa.Symmetric: Assume aRb, where a, b ∈ Z.
Then, a− b = 4k for somek ∈ Z. Then, b− a = 4(−k), and since −k ∈ Z, it follows thatb− a ∈ H. Therefore, bRa.
Another Example
Example
Let H = {4k | k ∈ Z}.A relation R is defined on Z by aRb ifa− b ∈ H. Show that R is an equivalence relation.
Proof.Reflexive: Let a ∈ Z. Since a− a = 0 = 4 · 0, it follows that aRa.Symmetric: Assume aRb, where a, b ∈ Z. Then, a− b = 4k for somek ∈ Z.
Then, b− a = 4(−k), and since −k ∈ Z, it follows thatb− a ∈ H. Therefore, bRa.
Another Example
Example
Let H = {4k | k ∈ Z}.A relation R is defined on Z by aRb ifa− b ∈ H. Show that R is an equivalence relation.
Proof.Reflexive: Let a ∈ Z. Since a− a = 0 = 4 · 0, it follows that aRa.Symmetric: Assume aRb, where a, b ∈ Z. Then, a− b = 4k for somek ∈ Z. Then, b− a = 4(−k), and since −k ∈ Z, it follows thatb− a ∈ H. Therefore, bRa.
Another Example
Proof.Transitive:
Assume aRb and bRc for a, b, c ∈ Z. Then a− b ∈ H andb− c ∈ H. Then, there exists integers k and l such that a− b = 4kand b− c = 4l. Therefore,
a− c = (a− b) + (b− c)
= 4k + 4l
= 4(k + l)
Since k + l ∈ Z, it follows that a− c ∈ H and so aRc.
Another Example
Proof.Transitive: Assume aRb and bRc for a, b, c ∈ Z.
Then a− b ∈ H andb− c ∈ H. Then, there exists integers k and l such that a− b = 4kand b− c = 4l. Therefore,
a− c = (a− b) + (b− c)
= 4k + 4l
= 4(k + l)
Since k + l ∈ Z, it follows that a− c ∈ H and so aRc.
Another Example
Proof.Transitive: Assume aRb and bRc for a, b, c ∈ Z. Then a− b ∈ H andb− c ∈ H.
Then, there exists integers k and l such that a− b = 4kand b− c = 4l. Therefore,
a− c = (a− b) + (b− c)
= 4k + 4l
= 4(k + l)
Since k + l ∈ Z, it follows that a− c ∈ H and so aRc.
Another Example
Proof.Transitive: Assume aRb and bRc for a, b, c ∈ Z. Then a− b ∈ H andb− c ∈ H. Then, there exists integers k and l such that a− b = 4kand b− c = 4l.
Therefore,
a− c = (a− b) + (b− c)
= 4k + 4l
= 4(k + l)
Since k + l ∈ Z, it follows that a− c ∈ H and so aRc.
Another Example
Proof.Transitive: Assume aRb and bRc for a, b, c ∈ Z. Then a− b ∈ H andb− c ∈ H. Then, there exists integers k and l such that a− b = 4kand b− c = 4l. Therefore,
a− c = (a− b) + (b− c)
= 4k + 4l
= 4(k + l)
Since k + l ∈ Z, it follows that a− c ∈ H and so aRc.
Another Example
Proof.Transitive: Assume aRb and bRc for a, b, c ∈ Z. Then a− b ∈ H andb− c ∈ H. Then, there exists integers k and l such that a− b = 4kand b− c = 4l. Therefore,
a− c = (a− b) + (b− c)
= 4k + 4l
= 4(k + l)
Since k + l ∈ Z, it follows that a− c ∈ H and so aRc.
Another Example
Proof.Transitive: Assume aRb and bRc for a, b, c ∈ Z. Then a− b ∈ H andb− c ∈ H. Then, there exists integers k and l such that a− b = 4kand b− c = 4l. Therefore,
a− c = (a− b) + (b− c)
= 4k + 4l
= 4(k + l)
Since k + l ∈ Z, it follows that a− c ∈ H and so aRc.
Another Example
Proof.Transitive: Assume aRb and bRc for a, b, c ∈ Z. Then a− b ∈ H andb− c ∈ H. Then, there exists integers k and l such that a− b = 4kand b− c = 4l. Therefore,
a− c = (a− b) + (b− c)
= 4k + 4l
= 4(k + l)
Since k + l ∈ Z, it follows that a− c ∈ H and so aRc.
And Another Example
Example
Determine if R is an equivalence relation if (x, y) ∈ R if 3 dividesx + 2y.
Proof.Reflexive: Let x ∈ Z. Then, x + 2x = 3x and 3 | 3x. So, xRx.Symmetric: Suppose x, y ∈ Z such that xRy. Then, x + 2y = 3k forsome integer k. That is, x = 3k − 2y. Consider
y + 2x = y + 2(3k − 2y)
= 6k − 3y
= 3(2k − y)
Since 2k − y ∈ Z, 3 | (y + 2x) and so yRx.
And Another Example
Example
Determine if R is an equivalence relation if (x, y) ∈ R if 3 dividesx + 2y.
Proof.Reflexive:
Let x ∈ Z. Then, x + 2x = 3x and 3 | 3x. So, xRx.Symmetric: Suppose x, y ∈ Z such that xRy. Then, x + 2y = 3k forsome integer k. That is, x = 3k − 2y. Consider
y + 2x = y + 2(3k − 2y)
= 6k − 3y
= 3(2k − y)
Since 2k − y ∈ Z, 3 | (y + 2x) and so yRx.
And Another Example
Example
Determine if R is an equivalence relation if (x, y) ∈ R if 3 dividesx + 2y.
Proof.Reflexive: Let x ∈ Z. Then, x + 2x = 3x and 3 | 3x. So, xRx.
Symmetric: Suppose x, y ∈ Z such that xRy. Then, x + 2y = 3k forsome integer k. That is, x = 3k − 2y. Consider
y + 2x = y + 2(3k − 2y)
= 6k − 3y
= 3(2k − y)
Since 2k − y ∈ Z, 3 | (y + 2x) and so yRx.
And Another Example
Example
Determine if R is an equivalence relation if (x, y) ∈ R if 3 dividesx + 2y.
Proof.Reflexive: Let x ∈ Z. Then, x + 2x = 3x and 3 | 3x. So, xRx.Symmetric: Suppose x, y ∈ Z such that xRy. Then, x + 2y = 3k forsome integer k.
That is, x = 3k − 2y. Consider
y + 2x = y + 2(3k − 2y)
= 6k − 3y
= 3(2k − y)
Since 2k − y ∈ Z, 3 | (y + 2x) and so yRx.
And Another Example
Example
Determine if R is an equivalence relation if (x, y) ∈ R if 3 dividesx + 2y.
Proof.Reflexive: Let x ∈ Z. Then, x + 2x = 3x and 3 | 3x. So, xRx.Symmetric: Suppose x, y ∈ Z such that xRy. Then, x + 2y = 3k forsome integer k. That is, x = 3k − 2y.
Consider
y + 2x = y + 2(3k − 2y)
= 6k − 3y
= 3(2k − y)
Since 2k − y ∈ Z, 3 | (y + 2x) and so yRx.
And Another Example
Example
Determine if R is an equivalence relation if (x, y) ∈ R if 3 dividesx + 2y.
Proof.Reflexive: Let x ∈ Z. Then, x + 2x = 3x and 3 | 3x. So, xRx.Symmetric: Suppose x, y ∈ Z such that xRy. Then, x + 2y = 3k forsome integer k. That is, x = 3k − 2y. Consider
y + 2x = y + 2(3k − 2y)
= 6k − 3y
= 3(2k − y)
Since 2k − y ∈ Z, 3 | (y + 2x) and so yRx.
And Another Example
Example
Determine if R is an equivalence relation if (x, y) ∈ R if 3 dividesx + 2y.
Proof.Reflexive: Let x ∈ Z. Then, x + 2x = 3x and 3 | 3x. So, xRx.Symmetric: Suppose x, y ∈ Z such that xRy. Then, x + 2y = 3k forsome integer k. That is, x = 3k − 2y. Consider
y + 2x = y + 2(3k − 2y)
= 6k − 3y
= 3(2k − y)
Since 2k − y ∈ Z, 3 | (y + 2x) and so yRx.
And Another Example
Example
Determine if R is an equivalence relation if (x, y) ∈ R if 3 dividesx + 2y.
Proof.Reflexive: Let x ∈ Z. Then, x + 2x = 3x and 3 | 3x. So, xRx.Symmetric: Suppose x, y ∈ Z such that xRy. Then, x + 2y = 3k forsome integer k. That is, x = 3k − 2y. Consider
y + 2x = y + 2(3k − 2y)
= 6k − 3y
= 3(2k − y)
Since 2k − y ∈ Z, 3 | (y + 2x) and so yRx.
And Another Example
Example
Determine if R is an equivalence relation if (x, y) ∈ R if 3 dividesx + 2y.
Proof.Reflexive: Let x ∈ Z. Then, x + 2x = 3x and 3 | 3x. So, xRx.Symmetric: Suppose x, y ∈ Z such that xRy. Then, x + 2y = 3k forsome integer k. That is, x = 3k − 2y. Consider
y + 2x = y + 2(3k − 2y)
= 6k − 3y
= 3(2k − y)
Since 2k − y ∈ Z, 3 | (y + 2x) and so yRx.
And Another Example
Proof.Transitive:
Suppose for x, y, z ∈ Z, we have that xRy and yRz. Then,for some k, l ∈ Z, x + 2y = 3k and y + 2z = 3l. That is, x = 3k − 2yand 2z = 3l− y. Consider
x + 2z = (3k − 2y) + (3l− y)
= 3k + 3l− 3y
= 3(k + l− y)
Since k + l− y ∈ Z, 3 | (x + 2z) and so xRz.
And Another Example
Proof.Transitive: Suppose for x, y, z ∈ Z, we have that xRy and yRz.
Then,for some k, l ∈ Z, x + 2y = 3k and y + 2z = 3l. That is, x = 3k − 2yand 2z = 3l− y. Consider
x + 2z = (3k − 2y) + (3l− y)
= 3k + 3l− 3y
= 3(k + l− y)
Since k + l− y ∈ Z, 3 | (x + 2z) and so xRz.
And Another Example
Proof.Transitive: Suppose for x, y, z ∈ Z, we have that xRy and yRz. Then,for some k, l ∈ Z, x + 2y = 3k and y + 2z = 3l.
That is, x = 3k − 2yand 2z = 3l− y. Consider
x + 2z = (3k − 2y) + (3l− y)
= 3k + 3l− 3y
= 3(k + l− y)
Since k + l− y ∈ Z, 3 | (x + 2z) and so xRz.
And Another Example
Proof.Transitive: Suppose for x, y, z ∈ Z, we have that xRy and yRz. Then,for some k, l ∈ Z, x + 2y = 3k and y + 2z = 3l. That is, x = 3k − 2yand 2z = 3l− y.
Consider
x + 2z = (3k − 2y) + (3l− y)
= 3k + 3l− 3y
= 3(k + l− y)
Since k + l− y ∈ Z, 3 | (x + 2z) and so xRz.
And Another Example
Proof.Transitive: Suppose for x, y, z ∈ Z, we have that xRy and yRz. Then,for some k, l ∈ Z, x + 2y = 3k and y + 2z = 3l. That is, x = 3k − 2yand 2z = 3l− y. Consider
x + 2z = (3k − 2y) + (3l− y)
= 3k + 3l− 3y
= 3(k + l− y)
Since k + l− y ∈ Z, 3 | (x + 2z) and so xRz.
And Another Example
Proof.Transitive: Suppose for x, y, z ∈ Z, we have that xRy and yRz. Then,for some k, l ∈ Z, x + 2y = 3k and y + 2z = 3l. That is, x = 3k − 2yand 2z = 3l− y. Consider
x + 2z = (3k − 2y) + (3l− y)
= 3k + 3l− 3y
= 3(k + l− y)
Since k + l− y ∈ Z, 3 | (x + 2z) and so xRz.
And Another Example
Proof.Transitive: Suppose for x, y, z ∈ Z, we have that xRy and yRz. Then,for some k, l ∈ Z, x + 2y = 3k and y + 2z = 3l. That is, x = 3k − 2yand 2z = 3l− y. Consider
x + 2z = (3k − 2y) + (3l− y)
= 3k + 3l− 3y
= 3(k + l− y)
Since k + l− y ∈ Z, 3 | (x + 2z) and so xRz.
And Another Example
Proof.Transitive: Suppose for x, y, z ∈ Z, we have that xRy and yRz. Then,for some k, l ∈ Z, x + 2y = 3k and y + 2z = 3l. That is, x = 3k − 2yand 2z = 3l− y. Consider
x + 2z = (3k − 2y) + (3l− y)
= 3k + 3l− 3y
= 3(k + l− y)
Since k + l− y ∈ Z, 3 | (x + 2z) and so xRz.
Congruences and Equivalence Relations
Theorem
Let m ∈ Z+.
(a) For every x ∈ Z, x ≡ x (mod m).
(b) For all x, y ∈ Z, if x ≡ y (mod m) then y ≡ x (mod m).
(c) For all x, y, z ∈ Z, if x ≡ y (mod m) and y ≡ z (mod m), thenx ≡ z (mod m).
Congruences and Equivalence Relations
Theorem
Let m ∈ Z+.
(a) For every x ∈ Z, x ≡ x (mod m).
(b) For all x, y ∈ Z, if x ≡ y (mod m) then y ≡ x (mod m).
(c) For all x, y, z ∈ Z, if x ≡ y (mod m) and y ≡ z (mod m), thenx ≡ z (mod m).
Congruences and Equivalence Relations
Theorem
Let m ∈ Z+.
(a) For every x ∈ Z, x ≡ x (mod m).
(b) For all x, y ∈ Z, if x ≡ y (mod m) then y ≡ x (mod m).
(c) For all x, y, z ∈ Z, if x ≡ y (mod m) and y ≡ z (mod m), thenx ≡ z (mod m).
The Proof
Proof.Proof of (a): x ∈ Z, Then m | 0,
Since x− x = 0, we have thatm | (x− x). That is, x ≡ x (mod m).
Proof of (b): Let x, y ∈ Z. Assume that x ≡ y (mod m). By definition,m | (x− y). Then, ∃k ∈ Z such that mk = x− y. But, m(−k) = y− x,so m | (y− x). That is, y ≡ x (mod m).
The Proof
Proof.Proof of (a): x ∈ Z, Then m | 0,Since x− x = 0, we have thatm | (x− x).
That is, x ≡ x (mod m).
Proof of (b): Let x, y ∈ Z. Assume that x ≡ y (mod m). By definition,m | (x− y). Then, ∃k ∈ Z such that mk = x− y. But, m(−k) = y− x,so m | (y− x). That is, y ≡ x (mod m).
The Proof
Proof.Proof of (a): x ∈ Z, Then m | 0,Since x− x = 0, we have thatm | (x− x). That is, x ≡ x (mod m).
Proof of (b): Let x, y ∈ Z. Assume that x ≡ y (mod m). By definition,m | (x− y). Then, ∃k ∈ Z such that mk = x− y. But, m(−k) = y− x,so m | (y− x). That is, y ≡ x (mod m).
The Proof
Proof.Proof of (a): x ∈ Z, Then m | 0,Since x− x = 0, we have thatm | (x− x). That is, x ≡ x (mod m).
Proof of (b): Let x, y ∈ Z. Assume that x ≡ y (mod m).
By definition,m | (x− y). Then, ∃k ∈ Z such that mk = x− y. But, m(−k) = y− x,so m | (y− x). That is, y ≡ x (mod m).
The Proof
Proof.Proof of (a): x ∈ Z, Then m | 0,Since x− x = 0, we have thatm | (x− x). That is, x ≡ x (mod m).
Proof of (b): Let x, y ∈ Z. Assume that x ≡ y (mod m). By definition,m | (x− y). Then, ∃k ∈ Z such that mk = x− y.
But, m(−k) = y− x,so m | (y− x). That is, y ≡ x (mod m).
The Proof
Proof.Proof of (a): x ∈ Z, Then m | 0,Since x− x = 0, we have thatm | (x− x). That is, x ≡ x (mod m).
Proof of (b): Let x, y ∈ Z. Assume that x ≡ y (mod m). By definition,m | (x− y). Then, ∃k ∈ Z such that mk = x− y. But, m(−k) = y− x,so m | (y− x).
That is, y ≡ x (mod m).
The Proof
Proof.Proof of (a): x ∈ Z, Then m | 0,Since x− x = 0, we have thatm | (x− x). That is, x ≡ x (mod m).
Proof of (b): Let x, y ∈ Z. Assume that x ≡ y (mod m). By definition,m | (x− y). Then, ∃k ∈ Z such that mk = x− y. But, m(−k) = y− x,so m | (y− x). That is, y ≡ x (mod m).
The Proof
Proof.Proof of (c): Let x, y, z ∈ Z. Assume that x ≡ y (mod m) andy ≡ z (mod m).
Since x ≡ y (mod m), ∃k ∈ Z such that mk = x− y.Since y ≡ z (mod m), ∃l ∈ Z such that ml = y− z.
Now,
x− z = (x− y) + (y− z)
= mk + ml
= m(k + l)
Since k + l ∈ Z, x− z = m(k + l) implies that m | (x− z). That is,x ≡ z (mod m).
The Proof
Proof.Proof of (c): Let x, y, z ∈ Z. Assume that x ≡ y (mod m) andy ≡ z (mod m). Since x ≡ y (mod m), ∃k ∈ Z such that mk = x− y.
Since y ≡ z (mod m), ∃l ∈ Z such that ml = y− z.
Now,
x− z = (x− y) + (y− z)
= mk + ml
= m(k + l)
Since k + l ∈ Z, x− z = m(k + l) implies that m | (x− z). That is,x ≡ z (mod m).
The Proof
Proof.Proof of (c): Let x, y, z ∈ Z. Assume that x ≡ y (mod m) andy ≡ z (mod m). Since x ≡ y (mod m), ∃k ∈ Z such that mk = x− y.Since y ≡ z (mod m), ∃l ∈ Z such that ml = y− z.
Now,
x− z = (x− y) + (y− z)
= mk + ml
= m(k + l)
Since k + l ∈ Z, x− z = m(k + l) implies that m | (x− z). That is,x ≡ z (mod m).
The Proof
Proof.Proof of (c): Let x, y, z ∈ Z. Assume that x ≡ y (mod m) andy ≡ z (mod m). Since x ≡ y (mod m), ∃k ∈ Z such that mk = x− y.Since y ≡ z (mod m), ∃l ∈ Z such that ml = y− z.
Now,
x− z = (x− y) + (y− z)
= mk + ml
= m(k + l)
Since k + l ∈ Z, x− z = m(k + l) implies that m | (x− z). That is,x ≡ z (mod m).
The Proof
Proof.Proof of (c): Let x, y, z ∈ Z. Assume that x ≡ y (mod m) andy ≡ z (mod m). Since x ≡ y (mod m), ∃k ∈ Z such that mk = x− y.Since y ≡ z (mod m), ∃l ∈ Z such that ml = y− z.
Now,
x− z = (x− y) + (y− z)
= mk + ml
= m(k + l)
Since k + l ∈ Z, x− z = m(k + l) implies that m | (x− z). That is,x ≡ z (mod m).
The Proof
Proof.Proof of (c): Let x, y, z ∈ Z. Assume that x ≡ y (mod m) andy ≡ z (mod m). Since x ≡ y (mod m), ∃k ∈ Z such that mk = x− y.Since y ≡ z (mod m), ∃l ∈ Z such that ml = y− z.
Now,
x− z = (x− y) + (y− z)
= mk + ml
= m(k + l)
Since k + l ∈ Z, x− z = m(k + l) implies that m | (x− z). That is,x ≡ z (mod m).
The Proof
Proof.Proof of (c): Let x, y, z ∈ Z. Assume that x ≡ y (mod m) andy ≡ z (mod m). Since x ≡ y (mod m), ∃k ∈ Z such that mk = x− y.Since y ≡ z (mod m), ∃l ∈ Z such that ml = y− z.
Now,
x− z = (x− y) + (y− z)
= mk + ml
= m(k + l)
Since k + l ∈ Z, x− z = m(k + l) implies that m | (x− z). That is,x ≡ z (mod m).
Example
Example
Let R be the relation defined on Z by aRb if 2a + b ≡ 0 (mod 3).Prove R is an equivalence relation.
Proof.Let x ∈ Z. Since 3 | (3x), it follows that 3x ≡ 0 (mod 3). So,2x + x ≡ 0 (mod 3). Thus, R is reflexive.
Example
Example
Let R be the relation defined on Z by aRb if 2a + b ≡ 0 (mod 3).Prove R is an equivalence relation.
Proof.Let x ∈ Z. Since 3 | (3x), it follows that 3x ≡ 0 (mod 3).
So,2x + x ≡ 0 (mod 3). Thus, R is reflexive.
Example
Example
Let R be the relation defined on Z by aRb if 2a + b ≡ 0 (mod 3).Prove R is an equivalence relation.
Proof.Let x ∈ Z. Since 3 | (3x), it follows that 3x ≡ 0 (mod 3). So,2x + x ≡ 0 (mod 3). Thus, R is reflexive.
Example
Proof.Next, we show R is symmetric. Assume xRy, where x, y ∈ Z.
Thus,2x + y ≡ 0 (mod 3) and so 3 | (2x + y), Therefore, 2x + y = 3r forsome r ∈ Z. Hence, y = 3r − 2x.So,
2y + x = 2(3r − 2x) + x
= 6r − 3x
= 3(2r + x)
Since 2r − x ∈ Z, 3 | (2y + x). So, 2y + x ≡ 0 (mod 3). Therefore,yRx and R is symmetric.
Example
Proof.Next, we show R is symmetric. Assume xRy, where x, y ∈ Z. Thus,2x + y ≡ 0 (mod 3) and so 3 | (2x + y),
Therefore, 2x + y = 3r forsome r ∈ Z. Hence, y = 3r − 2x.So,
2y + x = 2(3r − 2x) + x
= 6r − 3x
= 3(2r + x)
Since 2r − x ∈ Z, 3 | (2y + x). So, 2y + x ≡ 0 (mod 3). Therefore,yRx and R is symmetric.
Example
Proof.Next, we show R is symmetric. Assume xRy, where x, y ∈ Z. Thus,2x + y ≡ 0 (mod 3) and so 3 | (2x + y), Therefore, 2x + y = 3r forsome r ∈ Z.
Hence, y = 3r − 2x.So,
2y + x = 2(3r − 2x) + x
= 6r − 3x
= 3(2r + x)
Since 2r − x ∈ Z, 3 | (2y + x). So, 2y + x ≡ 0 (mod 3). Therefore,yRx and R is symmetric.
Example
Proof.Next, we show R is symmetric. Assume xRy, where x, y ∈ Z. Thus,2x + y ≡ 0 (mod 3) and so 3 | (2x + y), Therefore, 2x + y = 3r forsome r ∈ Z. Hence, y = 3r − 2x.So,
2y + x = 2(3r − 2x) + x
= 6r − 3x
= 3(2r + x)
Since 2r − x ∈ Z, 3 | (2y + x). So, 2y + x ≡ 0 (mod 3). Therefore,yRx and R is symmetric.
Example
Proof.Next, we show R is symmetric. Assume xRy, where x, y ∈ Z. Thus,2x + y ≡ 0 (mod 3) and so 3 | (2x + y), Therefore, 2x + y = 3r forsome r ∈ Z. Hence, y = 3r − 2x.So,
2y + x = 2(3r − 2x) + x
= 6r − 3x
= 3(2r + x)
Since 2r − x ∈ Z, 3 | (2y + x). So, 2y + x ≡ 0 (mod 3). Therefore,yRx and R is symmetric.
Example
Proof.Next, we show R is symmetric. Assume xRy, where x, y ∈ Z. Thus,2x + y ≡ 0 (mod 3) and so 3 | (2x + y), Therefore, 2x + y = 3r forsome r ∈ Z. Hence, y = 3r − 2x.So,
2y + x = 2(3r − 2x) + x
= 6r − 3x
= 3(2r + x)
Since 2r − x ∈ Z, 3 | (2y + x). So, 2y + x ≡ 0 (mod 3). Therefore,yRx and R is symmetric.
Example
Proof.Next, we show R is symmetric. Assume xRy, where x, y ∈ Z. Thus,2x + y ≡ 0 (mod 3) and so 3 | (2x + y), Therefore, 2x + y = 3r forsome r ∈ Z. Hence, y = 3r − 2x.So,
2y + x = 2(3r − 2x) + x
= 6r − 3x
= 3(2r + x)
Since 2r − x ∈ Z, 3 | (2y + x). So, 2y + x ≡ 0 (mod 3).
Therefore,yRx and R is symmetric.
Example
Proof.Next, we show R is symmetric. Assume xRy, where x, y ∈ Z. Thus,2x + y ≡ 0 (mod 3) and so 3 | (2x + y), Therefore, 2x + y = 3r forsome r ∈ Z. Hence, y = 3r − 2x.So,
2y + x = 2(3r − 2x) + x
= 6r − 3x
= 3(2r + x)
Since 2r − x ∈ Z, 3 | (2y + x). So, 2y + x ≡ 0 (mod 3). Therefore,yRx and R is symmetric.
Example
Proof.Finally, we show R is transitive. Assume that xRy and yRz, wherex, y, z ∈ Z.
Then 2x + y ≡ 0 (mod 3) and 2y + z ≡ 0 (mod 3). Thus,3 | (2x + y) and 3 | (2y + z). From this, it follows that 2x + y = 3rand 2y + z = 3s for some integers r and s. Adding these twoequations, we obtain
2x + 3y + z = 3r + 3s
so2x + z = 3r + 3s− 3y = 3(r + s− y)
Since r + s− y ∈ Z, 3 | (2x + z); so, 2x + z ≡ 0 (mod 3). Hence, xRzand so R is transitive.
Example
Proof.Finally, we show R is transitive. Assume that xRy and yRz, wherex, y, z ∈ Z. Then 2x + y ≡ 0 (mod 3) and 2y + z ≡ 0 (mod 3).
Thus,3 | (2x + y) and 3 | (2y + z). From this, it follows that 2x + y = 3rand 2y + z = 3s for some integers r and s. Adding these twoequations, we obtain
2x + 3y + z = 3r + 3s
so2x + z = 3r + 3s− 3y = 3(r + s− y)
Since r + s− y ∈ Z, 3 | (2x + z); so, 2x + z ≡ 0 (mod 3). Hence, xRzand so R is transitive.
Example
Proof.Finally, we show R is transitive. Assume that xRy and yRz, wherex, y, z ∈ Z. Then 2x + y ≡ 0 (mod 3) and 2y + z ≡ 0 (mod 3). Thus,3 | (2x + y) and 3 | (2y + z).
From this, it follows that 2x + y = 3rand 2y + z = 3s for some integers r and s. Adding these twoequations, we obtain
2x + 3y + z = 3r + 3s
so2x + z = 3r + 3s− 3y = 3(r + s− y)
Since r + s− y ∈ Z, 3 | (2x + z); so, 2x + z ≡ 0 (mod 3). Hence, xRzand so R is transitive.
Example
Proof.Finally, we show R is transitive. Assume that xRy and yRz, wherex, y, z ∈ Z. Then 2x + y ≡ 0 (mod 3) and 2y + z ≡ 0 (mod 3). Thus,3 | (2x + y) and 3 | (2y + z). From this, it follows that 2x + y = 3rand 2y + z = 3s for some integers r and s.
Adding these twoequations, we obtain
2x + 3y + z = 3r + 3s
so2x + z = 3r + 3s− 3y = 3(r + s− y)
Since r + s− y ∈ Z, 3 | (2x + z); so, 2x + z ≡ 0 (mod 3). Hence, xRzand so R is transitive.
Example
Proof.Finally, we show R is transitive. Assume that xRy and yRz, wherex, y, z ∈ Z. Then 2x + y ≡ 0 (mod 3) and 2y + z ≡ 0 (mod 3). Thus,3 | (2x + y) and 3 | (2y + z). From this, it follows that 2x + y = 3rand 2y + z = 3s for some integers r and s. Adding these twoequations, we obtain
2x + 3y + z = 3r + 3s
so2x + z = 3r + 3s− 3y = 3(r + s− y)
Since r + s− y ∈ Z, 3 | (2x + z); so, 2x + z ≡ 0 (mod 3). Hence, xRzand so R is transitive.
Example
Proof.Finally, we show R is transitive. Assume that xRy and yRz, wherex, y, z ∈ Z. Then 2x + y ≡ 0 (mod 3) and 2y + z ≡ 0 (mod 3). Thus,3 | (2x + y) and 3 | (2y + z). From this, it follows that 2x + y = 3rand 2y + z = 3s for some integers r and s. Adding these twoequations, we obtain
2x + 3y + z = 3r + 3s
so2x + z = 3r + 3s− 3y = 3(r + s− y)
Since r + s− y ∈ Z, 3 | (2x + z); so, 2x + z ≡ 0 (mod 3). Hence, xRzand so R is transitive.
Example
Proof.Finally, we show R is transitive. Assume that xRy and yRz, wherex, y, z ∈ Z. Then 2x + y ≡ 0 (mod 3) and 2y + z ≡ 0 (mod 3). Thus,3 | (2x + y) and 3 | (2y + z). From this, it follows that 2x + y = 3rand 2y + z = 3s for some integers r and s. Adding these twoequations, we obtain
2x + 3y + z = 3r + 3s
so2x + z = 3r + 3s− 3y = 3(r + s− y)
Since r + s− y ∈ Z, 3 | (2x + z); so, 2x + z ≡ 0 (mod 3). Hence, xRzand so R is transitive.